Area word problemsYou will accounter these area word problems often in math. Many of them will require familiarity with basic math, algebra skills, or a combination of both to solve the problems As I solve these area word problems, I will make an attempt to give you some problems solving skills Word problem #1: The area of a square is 4 inches. What is the length of a side? Important concept: Square. It means 4 equal sides. Area = s × s= 4 × 4 = 16 inches^{2} Word problem #2: A small square is located inside a bigger square. The length of one side of the small square is 3 inches and the length of one side of the big square is 7 inches What is the area of the region located outside the small square, but inside the big square? Important concept: Draw a picture and see the problem with your eyes. This is done below: The area that you are looking for is everything is red. So you need to remove the area of the small square from the area of the big square Area of big square = s × s = 7 × 7 = 49 inches^{2} Area of small square = s × s = 3 × 3 = 9 inches^{2} Area of the region in red = 49  9 = 40 inches^{2} Word problem #3: A classroom has a length of 20 feet and a width of 30 feet. The headmaster decided that tiles will look good in that class. If each tile has a length of 24 inches and a width of 36 inches, how many tiles are needed to fill the classroom? Important concept: Find area occupied by entire classroom Find area for one tile Decide which unit to use. In this case, we could use feet Area of classroom = length × width = 20 × 30 = 600 ft^{2} Before we get the area of each tile, convert the dimensions to feet as already stated Since 1 foot = 12 inches, 36 inches = 3 feet and 24 inches = 2 feet Area of each tile = length × width = 2 × 3 = 6 ft^{2} If one tile takes 6 ft^{2}, it takes 100 tiles to cover 600 ft^{2} (6 × 100 = 600) Word problem #4: Sometimes area word problems may require skills in algebra, such as factoring and solving quadratic equations A room whose area is 24 ft^{2} has a length that is 2 feet longer than the width. What are the dimensions of the room? Solution #1 Use of basic math skills and trial and error Pretend width = 1, then length = 3 ( 1 + 2) 1 × 3 is not equal to 24 and 3 is not close at all to 24. Try bigger numbers let width = 3, then length = 5 ( 3 + 2) 3 × 5 = 15. It is still not equal to 24 let width = 4, then length = 4 ( 4 + 2) 4 × 6 = 24. We are done! Solution #2 Use of algebra Let width = x, so length = x + 2 Area = length × width 24 = x × ( x + 2) 24 = x^{2} + 2x x^{2} + 2x = 24 (swap the left side with the right side) x^{2} + 2x  24 = 0 ( x + 6) × ( x  4 ) = 0 x = 6 and x = 4 So width = 4 and length = 4 + 2 = 6 As you can see area word problems can become very complicated as shows in the last problem. And the use of basic math skills may not be the best way to go to solve area word problems. Sometimes algebra is better because trial and error can take a long time! Have a question about these area word problems? send me a note. 




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